The minimum value

(a) 75

Given : $f(x)=x^{2}+\frac{250}{x}$

$\Rightarrow f^{\prime}(x)=2 x-\frac{250}{x^{2}}$

For a local maxima or a local minima, we must have'

$f^{\prime}(x)=0$

$\Rightarrow 2 x-\frac{250}{x^{2}}=0$

$\Rightarrow 2 x^{3}-250=0$

$\Rightarrow x^{3}=125$

$\Rightarrow x=5$

Now,

$f^{\prime \prime}(x)=2+\frac{500}{x^{3}}$

$\Rightarrow f^{\prime \prime}(5)=2+\frac{500}{5^{3}}=\frac{750}{125}=6>0$

So, $x=5$ is a local minima.

$\therefore f^{\prime}(x)_{\min }=5^{2}+\frac{250}{5}=\frac{375}{5}=75$